Difficulty: Introductory | Prerequisites: Resonance structures and delocalization (Ch. 1.5, Part 1), Lewis structures, electronegativity basics
Once you can draw resonance structures, the next question is: which one matters most? Formal charge is the bookkeeping tool that answers this. It tells you whether the electrons around each atom in a Lewis structure match what that atom "expects" as a free atom. By comparing formal charges across different resonance structures, you can rank them from most to least favourable and predict which contributes most to the resonance hybrid. This section builds directly on the resonance fundamentals and is essential for understanding acid-base strength, reaction mechanisms, and molecular stability in organic chemistry.
Formal charge = (valence electrons of the free atom) − (lone-pair electrons) − ½(bonding electrons). Use it to decide which resonance structure is most stable: the one with the fewest formal charges wins, and when charges do exist, negative charges should sit on the more electronegative atom. Equivalent structures (same formal charges on the same types of atoms) contribute equally to the hybrid.
Formal charge (FC)
A bookkeeping charge assigned to each atom in a Lewis structure. It compares the number of valence electrons the atom would have as a free, neutral atom against the number it "owns" in the structure. Think of it as asking: "Did this atom gain or lose electron ownership by being in this molecule?"
Electronegativity
A measure of how strongly an atom attracts shared electrons. In the context of formal charges, electronegative atoms (like O, N, F) are more "comfortable" bearing a negative formal charge, so structures that place negative charges on them are favoured.
Charge separation
When a resonance structure has both a positive and a negative formal charge on different atoms. In simple terms, charge separation costs energy, so structures with less of it are generally more stable.
Expanded octet
When a third-row element (like S or P) uses d-orbitals to accommodate more than eight electrons. This allows sulfur in SO₄²⁻ or phosphorus in PO₄³⁻ to form more bonds than second-row elements can, which affects formal charge distribution.
Equivalent resonance structures
Structures that have identical formal charge patterns and the same types of atoms bearing those charges. They contribute equally to the resonance hybrid, with no one structure being more favoured than any other.
Formal Charge = (number of valence electrons in free atom) − (number of lone-pair electrons) − ½(number of bonding electrons)
Alternatively: FC = V − L − B, where V = valence electrons of the free atom, L = lone-pair electrons on the atom, B = number of bonds to the atom.
Both formulations give the same result. The second is often faster because you can count bonds at a glance rather than tallying individual bonding electrons.
More covalent bonds = greater stability. The structure with the greatest number of covalent bonds is generally more stable because covalent bonds lower total energy.
Fewer formal charges = greater stability. A structure with no formal charges on any atom is preferred over one with formal charges.
Less charge separation = greater stability. When formal charges are unavoidable, the structure where the positive and negative charges are closer together (less separated) is more stable.
Negative charge on the more electronegative atom. If a structure places a negative formal charge on a more electronegative atom (like O rather than C), it is more stable.
Positive charge on the least electronegative atom. The converse of Rule 4: positive formal charges are better tolerated by less electronegative atoms.
Equivalent structures contribute equally. When two or more resonance structures have the same formal charge pattern on the same types of atoms, there is no difference in stability, and each contributes equally to the hybrid.
Valence electron count: S = 6; 4O = 24; plus 2 for the charge = 32 valence electrons.
Sulfur is a third-row element and can expand its octet, so structures with 5 or 6 bonds to sulfur are possible.
Six resonance structures can be drawn by varying which oxygen atoms carry double bonds to sulfur.
Because all four oxygen atoms are identical and equally electronegative, no single structure is favoured. All equivalent structures contribute equally to the hybrid (Rule 6).
Valence electron count: 2C = 8; 2O = 12; 3H = 3; plus 1 for the charge = 24 valence electrons.
Three resonance structures can be drawn for the carboxylate group.
The first and second structures, which place the negative formal charge on oxygen and have a C=O double bond, are the most favourable because they have the least formal charge overall (Rule 2).
The third structure, which places a positive charge on carbon and negative charges on both oxygens, is less favourable because of greater charge separation (Rule 3) and because carbon is less electronegative than oxygen (Rules 4 and 5).
Valence electron count: P = 5; H = 1; 3O = 18; plus 2 for the charge = 26 valence electrons.
Phosphorus, like sulfur, is a third-row element and can expand its octet.
Multiple resonance structures can be drawn by moving double bonds between P and different oxygen atoms.
The structure that places a positive formal charge on phosphorus (the least electronegative central atom) is less favourable than those distributing the charge across the oxygen atoms.
Valence electron count: C = 4; 2O = 12; H = 1; plus 1 for the charge = 18 valence electrons.
Two equivalent resonance structures can be drawn. In one, the double bond is between C and the first O; in the other, between C and the second O.
Both oxygens are identical, so the structures are equivalent and contribute equally (Rule 6). The hybrid has two identical C–O bonds of bond order 1.5.
Valence electron count: P = 5; 4O = 24; plus 3 for the charge = 32 valence electrons.
The resonance hybrid is drawn with dashed lines between P and each O, and the 3− charge is distributed equally across all four oxygen atoms.
The situation mirrors sulfate: all four oxygens are equivalent, so all structures contribute equally.
Formal charge analysis is how chemists decide which Lewis structure best represents a real molecule before they model its properties computationally. In drug design, for example, knowing where charge sits in a molecule helps predict how it will bind to an enzyme's active site, because electrostatic interactions between a drug and its target are driven by exactly the partial charges that formal charge bookkeeping approximates.
Students often confuse formal charge with actual charge (partial charge or oxidation state). Formal charge is a bookkeeping tool based on equal sharing of bonding electrons; it does not reflect the true electron density, which is influenced by electronegativity differences.
A common mistake is assuming the resonance structure with formal charges of zero is always the only contributor. It is the most favourable, but structures with small formal charges still contribute to the hybrid.
Students sometimes place a negative formal charge on the less electronegative atom and accept the structure as valid. It is technically valid as a Lewis structure, but it is a poor (less stable) contributor to the hybrid.
When working with third-row elements (S, P), students forget these atoms can expand their octet and draw structures that unnecessarily restrict them to four bonds.
⚠️ You will very likely be asked to calculate formal charges for every atom in a given Lewis structure. Drill the formula until it is instant: FC = V − L − B.
⚠️ Expect a question asking you to rank two or three resonance structures from most to least stable, using the six rules. Know the rules in order of priority.
⚠️ Sulfate (SO₄²⁻) and nitrate (NO₃⁻) are common exam molecules for drawing all possible resonance structures and identifying whether any is favoured.
⚠️ Be ready to explain why a structure with a negative charge on oxygen is favoured over one with a negative charge on carbon or nitrogen, using electronegativity.
Fill in the blank: Formal Charge = V − L − ______. Answer: B (the number of bonds to the atom).
True or false: A resonance structure with formal charges of zero on all atoms is always preferred over one with formal charges. Answer: True (assuming it otherwise obeys the Lewis structure rules).
True or false: Placing a negative formal charge on carbon rather than oxygen makes a resonance structure more stable. Answer: False. Negative charges are more stable on the more electronegative atom (oxygen).
Fill in the blank: When all resonance structures of an ion are equivalent, each one contributes ______ to the hybrid. Answer: equally.
True or false: Sulfur and phosphorus can only form four bonds, just like carbon. Answer: False. As third-row elements, they can expand their octet and form five or six bonds.
Q: Calculate the formal charge on the nitrogen atom in a Lewis structure of NO₃⁻ where nitrogen has one double bond and two single bonds to oxygen.
A: Nitrogen has 5 valence electrons as a free atom. In this structure it has 0 lone-pair electrons and 4 bonds. FC = 5 − 0 − 4 = +1.
Q: Two resonance structures of CH₃COO⁻ are drawn. Structure A places a negative formal charge on oxygen. Structure B places a negative charge on carbon and positive charge on oxygen. Which is more stable and why?
A: Structure A is more stable. It has fewer formal charges overall (Rule 2) and places the negative charge on the more electronegative atom, oxygen (Rule 4).
Q: The sulfate ion (SO₄²⁻) has six possible resonance structures. Is any single structure favoured over the others?
A: No. All four oxygen atoms are equivalent, so the resonance structures are equivalent. They contribute equally to the hybrid (Rule 6).
Q: What is the formal charge on each oxygen in a resonance structure of NO₃⁻ where that oxygen has a double bond to nitrogen and two lone pairs?
A: Oxygen has 6 valence electrons as a free atom. With 2 lone-pair electrons (4 electrons) and 2 bonds: FC = 6 − 4 − 2 = 0.
Q: A student draws a resonance structure for PO₄³⁻ in which phosphorus has only three bonds. Is this likely the best structure? Why or why not?
A: Probably not. Phosphorus is a third-row element and can form more than four bonds (expanded octet). Structures with more covalent bonds are generally more stable (Rule 1), so a structure with four or more P–O bonds would be favoured.
Formal charge analysis feeds directly into acid-base chemistry: the stability of a conjugate base depends on how well its negative charge is distributed, and formal charges tell you where the charge sits. It also connects to reaction mechanisms, where curved arrows push electrons toward atoms that can best accommodate the resulting formal charge. In later chapters on carbonyl chemistry and nucleophilic substitution, you will use formal charge reasoning almost every time you evaluate a reaction intermediate.
Formal charge, formal charge formula, FC = V − L − B, resonance stability, ranking resonance structures, charge separation, electronegativity and formal charge, expanded octet, third-row elements, sulfate resonance, acetate resonance, phosphate resonance, formate resonance, hydrogen phosphite, most stable resonance structure, Lewis structure formal charge, organic chemistry chapter 1.5, covalent bonds and stability, equivalent resonance contributors